Date | May 2013 | Marks available | 2 | Reference code | 13M.2.SL.TZ1.7 |
Level | Standard level | Paper | Paper 2 | Time zone | Time zone 1 |
Command term | Define | Question number | 7 | Adapted from | N/A |
Question
Define electric field strength.
The diagram shows a pair of horizontal metal plates. Electrons can be deflected vertically using an electric field between the plates.
(i) Label, on the diagram, the polarity of the metal plates which would cause an electron
positioned between the plates to accelerate upwards.
(ii) Draw the shape and direction of the electric field between the plates on the diagram.
(iii) Calculate the force on an electron between the plates when the electric field strength has a value of 2.5 × 103 NC–1.
The diagram shows two isolated electrons, X and Y, initially at rest in a vacuum. The initial separation of the electrons is 5.0 mm. The electrons subsequently move apart in the directions shown.
(i) Show that the initial electric force acting on each electron due to the other electron is approximately 9 × 10–24N.
(ii) Calculate the initial acceleration of one electron due to the force in (c)(i).
(iii) Discuss the motion of one electron after it begins to move.
(iv) The diagram shows Y as seen from X, at one instant. Y is moving into the plane of the paper. For this instant, draw on the diagram the shape and direction of the magnetic field produced by Y.
Markscheme
force per unit charge;
on a positive test charge / on a positive small charge;
(i) top plate positive and bottom negative (or +/- and ground);
(ii)
uniform (by eye) line spacing and edge effect, field lines touching both plates;
downward arrows (minimum of one and none upward);
(iii) F=2.5×103×1.6×10-19
4.0×10-16 (N);
Award [2] for a bald correct answer.
(i) use of \(F = \frac{{{{\left( {1.60 \times {{10}^{ - 19}}} \right)}^2}}}{{4\pi {\varepsilon _0}{{\left( {5.0 \times {{10}^{ - 3}}} \right)}^2}}}\) or \(F = \frac{{{{\left( {1.60 \times {{10}^{ - 19}}} \right)}^2}}}{{{{\left( {5.0 \times {{10}^{ - 3}}} \right)}^2}}} \times 8.99 \times {10^9}\);
9.2×10-24(N);
(ii) 1.0×107 (ms-2) (9.9×106 (ms-2) if 9×10-24 (N) used);
(iii) electron will continue to accelerate;
speed increases with acceleration;
acceleration reduces with separation;
when outside the field no further acceleration/constant speed;
any reference to accelerated charge radiating and losing (kinetic) energy;
(iv) minimum of two concentric circles centred on Y;
anti-clockwise;
Examiners report
(i) As this is worth two marks, candidates should see the signal that force per unit charge is unlikely to gain full marks; and so it proved. Although a mark was available for saying this there needed to be a reference to the charge being a positive test charge.
(i) G2 comments that the term ‘polarity’ was confusing to candidates proved to be unfounded and nearly all candidates marked in a positive and negative terminal – although the actual polarity was often incorrect.
(ii) With error carried forwards, the direction of the field was often correct but the drawing often was below an acceptable standard with line of force not bridging the plates, being very unevenly spaced and having no edge effect.
(iii) This calculation was almost invariably very well done.
(i) In another ‘show that’ question it was expected that candidates would use Coulombs law and the data value for the electronic charge to give a value of more than one digit; often this was not the case but otherwise this was generally well done
(ii) Most candidates used their value for the force (or 9 x 10-24 N) and the mass of the electron on the data sheet to calculate a correct value for the acceleration.
(iii) This was an unusual opportunity for candidates to use Newton’s laws and many did say that the acceleration would decrease with distance. Too often they incorrectly believed that this meant that the electron would slow down – it continues to accelerate but at an ever decreasing rate.
(iv) Clearly, this part represented a simplification of a complex situation but as set up was not beyond the skills of most of the candidates. The electron represents an instant in which a conventional current would leave the page and the field at this instant would be that of concentric circles with an anti-clockwise (counter-clockwise) direction. Many candidates did draw this but diagrams were too frequently hurriedly drawn and of a poor standard.